Question: $h(n) = -n^{3}-6n^{2}-2n+3(f(n))$ $f(n) = 4n^{2}-4n$ $ f(h(1)) = {?} $
Explanation: First, let's solve for the value of the inner function, $h(1)$ . Then we'll know what to plug into the outer function. $h(1) = -1^{3}-6(1^{2})+(-2)(1)+3(f(1))$ To solve for the value of $h$ , we need to solve for the value of $f(1)$ $f(1) = 4(1^{2})+(-4)(1)$ $f(1) = 0$ That means $h(1) = -1^{3}-6(1^{2})+(-2)(1)+(3)(0)$ $h(1) = -9$ Now we know that $h(1) = -9$ . Let's solve for $f(h(1))$ , which is $f(-9)$ $f(-9) = 4(-9)^{2}+(-4)(-9)$ $f(-9) = 360$